1. Jun 23, 2004

### needhelp

A person exerts a tangential force of 36.1 N on the rim of a disk-shaped merry-go-round of radius 2.74 m and mass 167 kg. If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 60.0°?

=> L=Iw => w = L/I where I=1/2mr^2
I just don't know where to plug in the tangential force and the angle.

2. Jun 24, 2004

### e(ho0n3

Let F be the tangential force, R the radius of the merry-go-round and M its mass. The torque is $\tau = FR = I\alpha$. You know F, R and I so you can solve for $\alpha$ and then use that to find the angular velocity. Pretty straight forward. Remember, physics is not about plugging numbers and spitting answers.

3. Jun 24, 2004

### needhelp

I calculated
a)alpha = T/I = 98.9/627 = 0.158 rad/s
b)theta = 1/2 (alpha) t^2 => t = sqrt(2 * 1.05 / 0.158) = 3.65s
c)w = (alpha) t = 0.158 rad/s^2 * 3.65 s = 0.577 rad/s

Please let me know if my figures are correct. Again, thank you.

4. Jun 24, 2004

### e(ho0n3

Instead of finding t, you could just use $\omega^2 = \omega_0^2 + 2\alpha\theta$. You can check your answer doing it this way.